Find if a crest is present in the index range [L, R] of the given array

Given an array arr[] of N distinct elements and an index range [L, R]. The task is to find whether a crest is present in that index range in the array or not. Any element arr[i] in the subarray arr[L...R] is called a crest if all the elements of the subarray arr[L...i] are strictly increasing and all elements of the subarray arr[i...R] are strictly decreasing.
Examples: 
 

Input: arr[] = {2, 1, 3, 5, 12, 11, 7, 9}, L = 2, R = 6 
Output: Yes 
Element 12 is a crest in the subarray {3, 5, 12, 11, 7}.
Input: arr[] = {2, 1, 3, 5, 12, 11, 7, 9}, L = 0, R = 2 
Output: No 
 

Approach: 
 

• Check whether in the given index range [L, R] there exists an element which satisfies the Property where arr[i - 1] ? arr[i] ? arr[i + 1].
• If any element in the given range satisfies the above property then the given range cannot contain a crest otherwise the crest is always possible.
• To find which element satisfies the above property, maintain an array present[] where present[i] will be 1 if arr[i - 1] ? arr[i] ? arr[i + 1] else present[i] will be 0.
• Now convert the present[] array to its cumulative sum where present[i] will now represent number of elements in the index range [0, i] that satisfy the property.
• For an index range [L, R] to contain a crest, present[L] must be equal to present[R - 1].

Below is the implementation of the above approach: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if the
// array contains a crest in
// the index range [L, R]
bool hasCrest(int arr[], int n, int L, int R)
{
    // To keep track of elements
    // which satisfy the Property
    int present[n] = { 0 };

    for (int i = 1; i <= n - 2; i++) {

        // Property is satisfied for
        // the current element
        if ((arr[i] <= arr[i + 1])
            && (arr[i] <= arr[i - 1])) {
            present[i] = 1;
        }
    }

    // Cumulative Sum
    for (int i = 1; i < n; i++) {
        present[i] += present[i - 1];
    }

    // If a crest is present in
    // the given index range
    if (present[L] == present[R - 1])
        return true;

    return false;
}

// Driver code
int main()
{
    int arr[] = { 2, 1, 3, 5, 12, 11, 7, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int L = 2;
    int R = 6;

    if (hasCrest(arr, N, L, R))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}

// Java implementation of the approach
import java.util.*;

class GFG
{
    
// Function that returns true if the
// array contains a crest in
// the index range [L, R]
static boolean hasCrest(int arr[], int n, 
                        int L, int R)
{
    // To keep track of elements
    // which satisfy the Property
    int []present = new int[n];
    for(int i = 0; i < n; i++)
    {
        present[i] = 0;
    }

    for (int i = 1; i <= n - 2; i++) 
    {

        // Property is satisfied for
        // the current element
        if ((arr[i] <= arr[i + 1]) && 
            (arr[i] <= arr[i - 1])) 
        {
            present[i] = 1;
        }
    }

    // Cumulative Sum
    for (int i = 1; i < n; i++) 
    {
        present[i] += present[i - 1];
    }

    // If a crest is present in
    // the given index range
    if (present[L] == present[R - 1])
        return true;

    return false;
}

// Driver code
public static void main(String args[])
{
    int arr[] = { 2, 1, 3, 5, 12, 11, 7, 9 };
    int N = arr.length;
    int L = 2;
    int R = 6;

    if (hasCrest(arr, N, L, R))
        System.out.println("Yes");
    else
        System.out.println("No");

}
}

// This code is contributed by Surendra_Gangwar

# Python3 implementation of the approach 

# Function that returns true if the 
# array contains a crest in 
# the index range [L, R] 
def hasCrest(arr, n, L, R) : 

    # To keep track of elements 
    # which satisfy the Property 
    present = [0] * n ; 

    for i in range(1, n - 2 + 1) :

        # Property is satisfied for 
        # the current element 
        if ((arr[i] <= arr[i + 1]) and 
            (arr[i] <= arr[i - 1])) :
            present[i] = 1; 

    # Cumulative Sum 
    for i in range(1, n) :
        present[i] += present[i - 1]; 

    # If a crest is present in 
    # the given index range 
    if (present[L] == present[R - 1]) :
        return True; 

    return False; 

# Driver code 
if __name__ == "__main__" : 

    arr = [ 2, 1, 3, 5, 12, 11, 7, 9 ]; 
    N = len(arr); 
    L = 2; 
    R = 6; 

    if (hasCrest(arr, N, L, R)) :
        print("Yes"); 
    else :
        print("No"); 

# This code is contributed by AnkitRai01

    
// C# implementation of the approach
using System;

class GFG
{
    
// Function that returns true if the
// array contains a crest in
// the index range [L, R]
static bool hasCrest(int []arr, int n, 
                        int L, int R)
{
    // To keep track of elements
    // which satisfy the Property
    int []present = new int[n];
    for(int i = 0; i < n; i++)
    {
        present[i] = 0;
    }

    for (int i = 1; i <= n - 2; i++) 
    {

        // Property is satisfied for
        // the current element
        if ((arr[i] <= arr[i + 1]) && 
            (arr[i] <= arr[i - 1])) 
        {
            present[i] = 1;
        }
    }

    // Cumulative Sum
    for (int i = 1; i < n; i++) 
    {
        present[i] += present[i - 1];
    }

    // If a crest is present in
    // the given index range
    if (present[L] == present[R - 1])
        return true;

    return false;
}

// Driver code
public static void Main(String []args)
{
    int []arr = { 2, 1, 3, 5, 12, 11, 7, 9 };
    int N = arr.Length;
    int L = 2;
    int R = 6;

    if (hasCrest(arr, N, L, R))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}

// This code is contributed by PrinciRaj1992

<script>

// JavaScript implementation of the approach 

// Function that returns true if the 
// array contains a crest in 
// the index range [L, R] 
function hasCrest(arr, n, L, R) 
{ 
    // To keep track of elements 
    // which satisfy the Property 
    let present = new Uint8Array(n); 

    for (let i = 1; i <= n - 2; i++) { 

        // Property is satisfied for 
        // the current element 
        if ((arr[i] <= arr[i + 1]) 
            && (arr[i] <= arr[i - 1])) { 
            present[i] = 1; 
        } 
    } 

    // Cumulative Sum 
    for (let i = 1; i < n; i++) { 
        present[i] += present[i - 1]; 
    } 

    // If a crest is present in 
    // the given index range 
    if (present[L] == present[R - 1]) 
        return true; 

    return false; 
} 

// Driver code 

    let arr = [ 2, 1, 3, 5, 12, 11, 7, 9 ]; 
    let N = arr.length; 
    let L = 2; 
    let R = 6; 

    if (hasCrest(arr, N, L, R)) 
        document.write("Yes"); 
    else
        document.write("No"); 


// This code is contributed by Surbhi Tyagi.

</script>

Yes

Time Complexity: O(n)

Auxiliary Space: O(n)